Note: As we are using the same technological type assignment, the procedures are the exact same as Lab 7.5.1,thus the same instructions below are the same as the previous instruction.

Procedure (Technological: wiki spaces):

1) To create a wiki for this project, first go to http://www.wikispaces.com/
2) Next click the Teacher button under Teachers & Students
3) Now click: Sign up and start your wiki
3) This will take you to a small form, fill in the form with your information. Under Make a Wiki, select yes, and give a name to it, in the next panel. Now pick your permissions, and the type of wiki, being created.
4) Once done you will be brought to your own wiki, you can then go to under the Members section to invite new members, or under the Manage Wiki section, press members, then go under permissions, then press allow membership requests, to allow people to request membership. In this panel, you can also edit the different permissions given to users. Note: Only creator's/admins can use this panel.
5) To create a page to do your project, you can either press projects under Wiki Home, or go to Pages and Files. On the top right in Pages and Files, you can press new page icon, which creates a new page.
6) To edit the new page you must be a member and can simply press Edit on the top right.
Note: currently anyone can look at this wiki, if they know the URL, but only the current group using this wiki can edit/create pages. This is due to the fact it costs money to make it fully private, and it will create a lot of hassle, trying to make sure you, the teacher, can see this site to mark it.

Procedure (Technological: commenting / looking at who did what aspect of assignment)

1) To find out who did what in this project, you can press the view revisions button. This shows every single person who worked on said page, and at what time. You can then select 2 of the revisions by pressing the select button under the Compare category. This will then show you the changes between both copies. Thus showing who added what to this page. To make it simpler, in the discussions section, we have specified who did what section.
2) To add your own comments for marking you can simply press edit, then press Comment, this will bring up a sticky note much like in our flex-book assignments.
3) To add a rubric, you can simply upload it under Pages and Files, then click upload.

GAS NOTES

The Nature of Gases
Goals for this section:
- Describe the assumptions of the kinetic theory as it applies to gases
- Interpret gas pressure in terms of kinetic theory
- Define the relationship between Kelvin temperature and average kinetic energy
- Terms to know: kinetic energy, kinetic theory, gas pressure, vacuum, atmospheric pressure, barometer, pascal (Pa), standard atmosphere (atm)

1. I. Kinetic Theory
   1. a. The word kinetic refers to motion
   2. b. The energy an object has because of its motion is called kinetic energy .
   3. c. According to the kinetic theory, all matter consists of tiny particles that are in constant motion.
   4. d. 3 Assumptions of kinetic theory:
      1. i.

         002

         The particles in a gas are considered to be small, hard spheres with an insignificant volume
         - This means only the volume between particles matters
         - Particle size doesn’t matter
      2. ii. The motion of the particles in a gas is rapid, constant, and random .
      3. iii. All collisions between particles in a gas are perfectly elastic- this means there are no attractions

- Particles are too far apart to be attracted to each other

1. II. Gas pressureresults from the force exerted by a gas per unit surface area of an object
   1. a. This means: gas pressure is the result of billions of collisions happening simultaneously because of rapidly moving particles in a gas with an object
   2. b. An empty space with no particles and no pressure is called a vacuum .
   3. c. Atmospheric pressure results from the collision of atoms and molecules in air with objects.
   4. d. A barometer is a device that is used to measure atmospheric pressure.

1. III. The SI unit of pressure is the pascal (Pa). However, kPa is usually used.
   1000 Pa = 1 kPa
   1. a. Standard pressure is 101.3 kPa or 1 atm . This is also the pressure at sea level.
   2. b.

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1. IV. Average Kinetic Energy
   1. a. The particles in any collection of atoms or molecules at a given temperature have a wide range of kinetic energies. Most of the particles have kinetic energies somewhere in the middle of this range
   2. b.

      006

   3. c. Absolute zero(0 K, or –273.15°C) is the temperature at which the motion of particles theoretically ceases.
      1. i. Particles would have no kinetic energy at absolute zero.
      2. ii. Absolute zero has never been produced in the laboratory.
   4. d. The Kelvin temperature of a substance is directly proportionalto the average kinetic energy of the particles of the substance.
      1. i. This means that :
         As temperature increases, KE increases As temperature decreases, KE decreases

Practice problem 1:

What pressure in kilopascals and in atmospheres does a gas exert at 385 mm Hg.
Given:
pressure= 385mm Hg
1 atm=101.3kPa
1 atm=760mm Hg
Required:
pressure= ? kPa
pressure= ? atm
For converting mm Hg à atm, the appropriate conversion factor is
1 atm/ 760 mm Hg
For converting mm Hg à kPa, the appropriate conversion factor is
101.3 kPa/760 mm Hg
355 mm Hg * 1 atm / 760 mm Hg = 0.467atm
355 mm Hg * 101.3 kPa / 760 mm Hg = 47.318 kPa
Because the first conversion factor is smaller than 1, and 355 is smaller than the denominator 760 as well, it makes sense that the answer is smaller than 1.
Because the second conversion factor has a numerator greater than one, the numerator when multiplied with amount of mm Hg will multiply up to a bigger number than 760 and thus it makes sense that the answer is greater than 1.

Notes on your own - Properties of Gases (there are 8 questions)
Goals for this section:
- Explain why gases are easier to compress than solids and liquids
- Describe the 3 factors that affect air pressure
- Explain how changes in these 3 factors result in changes in air pressure
- Terms to know: Compressibility

1. Compressibilityis a measurement of how much the volume of matter decreases under pressure. When a person collides with an inflated airbag, the compression of the gas absorbs the energy of the impact.
   1. Why does a gas compress more than a solid or liquid? The space between gas particles are much more than that of a liquid or solid, therefore when under pressure, the gas particles have the ability to move closer together.
   2. At room temperature, the distance between particles in an enclosed gas is about 10 times the diameter of a particle
2. There are three things that can impact gas pressure. What are they? (hint: think about the warnings on aerosol cans, how your tires behave in the summer vs. the winter, how you change the pressure in your tires, and what happens to a balloon if you squeeze it) The amount of gas, volume and temperature.
3. Four variables are generally used to describe a gas. The units and the symbol used to indicate the 4 variables are listed below. List what the symbol stands for.
   1. Pressure (P) in kilopascals
   2. Volume (V) in liters
   3. Temperature (T) in kelvins
   4. the number of moles (n).

1. Explain what is happening in the pictures below. The gas pressure increases until it exceeds the strength of an enclosed, rigid container, and then the container bursts.

1. As a substance or mixture is heated, do particles move faster or slower? As substance or mixture is heated, the average kinetic energy of the particles increases (therefore particles move faster).
2. Based on your answer to number 5, will the particles hit the walls of their containers with relatively high energy or with relatively low energy when temperatures are high? When the temperatures are high the faster-moving particles strike the walls of their container with more energy.
3. Explain what is happening in the pictures below: The Kelvin temperature of the enclosed gas doubles causing the pressure of the enclosed gas to double.

1. Explain what is happening in the pictures below: The volume of the container is halved causing the pressure of the gas exerted to be doubled.

14-01-15a

14-01-15b

Gas Laws: Charles’, Boyle’s, Gay-Lussac’s, Combined
Goals for this section:
- Describe the relationships among temperature, volume, and pressure of gases
- Use Boyle’s, Charles’, Gay-Lussac’s, and the combined gas law to solve problems involving gases
- Terms to know: Boyle’s Law, Charles’ Law, Gay-Lussac’s Law, combined gas law

1. Boyle’s Law- Pressure and Volume
2. a. Pressure increases as you fill a tire with air because: As you increase the # of gas particles the # of collisions between particles and the walls of the container also increases
3. b. Reducing volume causes an increase in pressure because: reducing the volume causes the number of collisions with the container to increase, increasing pressure
4. c. Boyle’s Law- For a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure.
   1. a. As one goes up, the other goes down.
   2. b. P x V = K Pressure times volume is constant provided the temperature and amount of gas do not change.

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EXAMPLE:

002

A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant? Given: V1 = 4L
P1 = 205kPa V2 = 12.0L Required: P2 Analyze: I will need to use Boyle’s law to solve the problem and find P2. The formula needed is P1 x V1 = P2 x V2. Solve: P2 = (P1 x V1)/ V2
P2 = (205kPa x 4L)/ 12L (The L will cancel out)
P2 = 68.33kPa
Paraphrase:
Therefore the pressure of the container will be at 68.33kPa, if the temperature is constant.
Justify: I know the answer I got is correct, due to the fact if the volume increases, then the pressure will have to decrease to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Boyle’s law, when one goes up, the other must go down, which occurs here.

Charles’ Law- Temperature and volume

1. A. Relationship between temperature and volume can only be studied for a limited range of temperatures because: At low temperature, gases condense
2. B. Charles’s Law: The volume of a fixed mass of gas is directly proportional to its KELVIN temperature if the pressure is constant.

1) as one variable goes up, the other goes up.
2) V/T = K at constant mass and pressure.

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EXAMPLE PROBLEM:

004

Exactly 5.00 L of air at -50.0oC is warmed to 100.0 oC. What is the new volume if the pressure remains constant? Given: V1 = 5L
T1 = -50.0oC T­2 = 100.0 oC Required: V2 Analyze: I will need to use Charles’s law to solve the problem and find V2. The formula needed is V1/ T1 = V2/ T2. Solve: V2 = (V1 x T2)/ T1. To convert from Celsius to Kelvin, I need to use the formula K = C +273.
V2 = (5L x (100.0 oC+273K/1C))/ (-50.0oC+273K/1C) (The C and K will cancel out)
V2 = 8.36L
Paraphrase:
Therefore the new volume will be 8.36L, if the pressure is constant.
Justify: I know the answer I got is correct, due to the fact if the temperature increases, then the volume will have to decrease to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Charles’s law, when one goes up, the other must go up, which occurs here.

1. Gay-Lussac’s Law- Temperature and Pressure
2. A. When temperature increases, particles move faster and hit the walls of the container with greater energy, increasing pressure
3. B. Gay- Lussac’s law- Pressure is directly proportional to the KELVIN temperature if the volume and mass remain constant.
4. C. When a gas is heated at constant volume, the pressure increases.
5. D. When one variable increases, the other variable increases

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EXAMPLE PROBLEM: The pressure in a car tire is 198 kPa at 27oC. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant. Given: T1 = 27oC + (273K/1C)
P1 = 198kPa P2 = 225kPa Required: T2 Analyze: I will need to use Gay-Lussac’s law to solve the problem and find T2. The formula needed is P1/ T1 = P2/ T2. Solve: T2 = (T1 x P2)/P1
T2 = (300K x 225kPa)/(198kPa) (The kPa will cancel out)
T2 = 340.91K
Paraphrase:
Therefore temperature of the air in the tire is at 340.91K, if the volume is constant.
Justify: I know the answer I got is correct, due to the fact if the pressure increases, then the temperature will have to increase to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Gay-Lussac’s law, when one goes up, the other must go up, which occurs here.

1. Combined gas law
2. A. The combined gas law allows you to do calculations for situations in which only the amount of gas is constant

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EXAMPLE PROBLEM: A 5.00 L air sample has a pressure of 107 kPa at a temperature of -50.0oC. IF the temperature is raised to 102oC and the volume expands to 7.00 L, what will the new pressure be? Given:T1 = -50.0oC + (273K/1C)
P1 = 107kPa V1 = 5L T2 = 102oC + (273K/1C) V2 = 7L Required: P2 Analyze: I will need to use Combined Gas law to solve the problem and find P2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: P2 = (P1 x T2 x V1)/(T1 x V2)
P2 = (107kPax 375K x 5L)/(223K x 7L) (The L and K will cancel out)
P2 = 128.52kPa
Paraphrase:
Therefore the new pressure will be at 128.52kPa.
Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

1. Tips for Gas Law Problems

1) Determine which gas law you need
n Pressure and volume = Boyle’s
n Temperature and volume = Charles’
n Temperature and Pressure = Gay-Lussac’s
n Temperature, pressure, and volume = Combined
2) Identify your variables. Be sure you put the proper numbers together
3) Change all variables into the correct units
n Temperature must be Kelvin
n Pressure units must match
n Volume units must match
4) Put numbers into the gas law equation and solve


Ideal gas Law
Goals for this section:
- Calculate the value of an unknown variable using the ideal gas law
- Compare and contrast real and ideal gases
- Terms to know: ideal gas law, ideal gas constant
VI. Ideal Gas Law

1. A. Ideal gas- A gas that conforms to the gas laws.
   1. a. Particles exhibit no attractive or repulsive forces .
   2. b. Volume of particles is assumed to be negligible compared to the total volume occupied by the gas. This means you can ignore the volume of the particles
2. B. Real gases do not behave ideally when they have high pressures or low temperatures
   1. a. Under these conditions, gases tend to condense
3. C. When is a gas ideal?
   1. a. When it has large volume and low pressure and/or high temperature
   2. b. Under these circumstances, attractive forces are not important
4. D. ideal gas law:

   e0426-01

   .

P = pressure (must be in units of kPa)
V = volume (must be in units of L)
n = moles (may need to convert grams or molecules to moles first)
R = 8.31 LkPa/(Kmol)
T = temperature (must be in units of K)

1. E. Example Problem: A child’s lungs can hold 2.20 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a body temperature of 37oC? Use a molar mass of 29 grams for air.

Given: P= 102kPa, V=2.20L, T =37°C, R = 8.31LkPa/(Kmol), molar mass air = 29.0g
Required: g of air
Analyze: First the temperature should be in the units of K, and then the moles should be found.
Solve:
0°C = 273K
K of 37°C= 273K + 37°C
= 310K
n=P x V/ R x T
n= (102kPa) x (2.20L) / 8.31 LkPa/(Kmol) x 310K = 0.0871 mol air
0.0871 mol air x 29g air / 1 mol air = 2.52g air
Paraphrase: Therefore her lungs hold 2.52g of air.
Justify: This answer makes sense because the moles of air is small, therefore the grams will be very small too.

Partial Pressures, Diffusion, Effusion
Goals for this section:
- Relate the total pressure of a mixture of gases to the partial pressures of the component gases
- Explain how the molar mass of a gas affects the rates at which the gas diffuses and effuses
- Compare and contrast effusion and diffusion
- Terms to know: partial pressure, Dalton’s law of partial pressures, diffusion, effusion, Graham’s law of effusion/diffusion

1. VII. Partial Pressures, Diffusion, Effusion
   1. a. Partial Pressure:
      1. i. The contribution each gas in a mixture makes to the total pressure is called the partial pressure exerted by that gas
         1. 1. Dalton’s law of partial pressures; In a mixture of gases, the total pressure is the sum of the partial pressures of the gases
         2. 2. Provided the temperature and volume remain the same
   2. b. Diffusion is the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout
   3. c. During effusion, a gas escapes through a tiny hole in its container .
   4. d. Gases of lower mass diffuse and effuse faster than gases of higher molar mass.
   5. e. Graham’s law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass. This law can also be applied to the diffusion of gases.

Presentation 5
32. A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.9 kPa. If P 02 = 6.6 kPa and P N2=23.0 kPa, what is P CO2
PN2=23 kPa
P02=6.6 kPa
Ptotal=32.9 kPa
PCO2= ? kPa
Ptotal= PN2+ P02+ PCO2
PCO2= Ptotal - PN2- P02
= 32.9 kPa – 23 kPa – 6.6 kPa
= 3.3 kPa
The partial pressure of carbon dioxide should be smaller than that of nitrogen because Ptotal is only 32.9 kPa. The partial pressure for oxygen is small too so the answer 3.3 kPa seems reasonable.

If I place 3 moles of N2 and 4 moles of O2 in a 35 L container at a temperature of 250 C, what will the pressure of the resulting mixture of gases be?

Moles N2 and O2 = 3 mol + 4 mol = 7 mol
Volume = 35 L
Temperature = 250C = (25+ 273.15) K = 298.15 K
R = 8.31 LkPa/(Kmol)
P = ? kPa
P=nRT/V
= 0.007 Kmol* * 298.15 K *
= 0.496 kPa
Because the amount of Kmol in this problem is only 0.007 and the answer is in kPa, it makes sense that the answer will be smaller than one. Thus the answer 0.496 kPa seems reasonable.

Compare the effusion rates of nitrogen and hydrogen gases.
Molar mass H2­­=1.01g
Molar mass N2­­=14.01g

So from the same hole hydrogen will escape 3.72 times faster than nitrogen. This makes sense because at the same temperature both gases possess the same kinetic energy but the mass of nitrogen is higher, and thus the velocity of hydrogen will be higher as it has less mass.

GAS WORKSHEETS

Practice converting pressure units and Kinetic Theory

1. 1. What are the 3 assumptions of the kinetic theory? The first assumpmtion of the kinetic theory is that the particles in a gas are considered to be small, hard spheres with an insignificant volume. The second assumption is that the motion of the particles in a gas is rapid, constant, and random. The third assumption is that all collisions between particles in a gas are perfectly elastic.
2. 2. An empty space with no particles and no pressure is called a vacuum
3. 3. What is the source of pressure? The source of gas pressure is the force exerted by a gas per unit surface area of an object. In other words the source of gas pressure is the force exerted by a gas when billions of rapidly moving particles collide with each other
4. 4. Describe the particle motion at absolute zero: The particle motion at absolute zero theoretically completely stops and it is achieved at 0 K or –273.15°C.
5. 5. If a gas heats up, do particles move faster or slower? As gas heats up, the average kinetic energy of the particles in the gas increases causing faster movement in gas particles. Assuming the container is rigid, how does an increase in temperature impact pressure? The increase in the temperature of an enclosed gas causes an increase in its pressure, and if the temperature doubles, the pressure doubles simultaneously. But if the gas pressure increases until it exceeds the strength of an enclosed, rigid container, then the container will burst.
6. 6. If a gas is compressed, do particles hit the walls of the container more or less often? How does this impact pressure(assuming the container is rigid? When the volume of the container is reduced, the pressure of the particles will increase, and with the increase in pressure there is an increase temperature, therefore the particles tend to move faster. And faster-moving particles strike the walls of their container with more energy and more frequently because they have less room to cover since the volume is decreased and their speed has been increased too.
7. 7. If more gas particles are added to a container, what happens to the pressure of the container? Why? Adding more gas particles to a container is similar to compressing the gas particles, because when compressing, more particles are being present in a smaller volume. Similarly, adding gas particles will cause more gas particles to be present within a given volume thus increasing the pressure.
8. 8. For each of the following, convert from the units given to the desired units using dimensional analysis. SHOW YOUR WORK if you wish to receive credit. Include all units and box or highlight your answers
   1. a. 130.5 kPa = ? atm Given: pressure=130.5 kPa Required: pressure=?atm For converting kPa à atm, the appropriate conversion factor is 1 atm/ 101.3 kPa pressure= 130.5kPa*1atm/101.3kPa=1.29atm Therefore 130.5 kPa is equal to 1.29 atm The answer 1.29 atm makes sense because at 1 atm the pressure is 101.3 kPa and since 130.5 kPa is larger than 101.3 kPa it makes sense that the pressure is also greater than 1 atm
   2. b. 45 kPa = ? mmHg Given: pressure=45 kPa Required: pressure=?mmHg For converting kPa à mm Hg, the appropriate conversion factor is 760 mm Hg/ 101.3 kPa pressure= 45 kPa*760mmHg/101.3kPa=337.61 mm Hg Therefore 45 kPa is equal to 337.61 mm Hg Because the conversion factor is greater than 1, it makes sense that the answer will be greater than 1 and the answer 337.61 mm Hg seems reasonable.

1. c. 1.2 atm = ? kPa Given: pressure=1.2 atm Required: pressure=?kPa For converting atm à kPa, the appropriate conversion factor is 101.3 kPa/1 atm pressure= 1.2atm*101.3kPa/1atm= 121.56kPa Therefore 1.2 atm is equal to 121.56 kPa The answer 121.56kPa makes sense because at 1 atm the pressure is 101.3 kPa and since 1.2 atm is larger than 1 atm it makes sense that the pressure is also greater than 101.3 kPa
2. d. 0.76 atm = ? mmHg Given: pressure=0.76 atm Required: pressure=?mmHg For converting atm à mmHg, the appropriate conversion factor is 760mmHg/1atm pressure= 0.76*760mmHg/1atm= 577.6 mm Hg Therefore 0.76 atm is equal to 577.60 mm Hg Because the conversion factor is a lot greater than one it makes sense that the answer will be a lot greater than one and thus 577.6 mm Hg for 0.76 atm makes sense
3. e. 770 mmHg = ? kPa Given: pressure=770 mmHg Required: pressure=?kPa For converting mm Hg à kPa, the appropriate conversion factor is 760 mm Hg/101.3 kPa pressure= 770 mm Hg*101.3kPa/760 mm Hg= 102.63kPa Therefore 770 mm Hg is equal to 102.63 kPa 760 mm Hg is equivalent to 101.3 kPa, and since 770 mm Hg is only a 10 units above 760 mmHg it makes sense that the pressure in kPa will only be a little greater than 101.3 kPa and thus the answer 102.63 kPa seems reasonable
4. f. 25 mmHg = ? atm Given: pressure=25mmHg Required: pressure=?atm For converting mm Hg à atm, the appropriate conversion factor is 1 atm/ 760 mmHg pressure= 25mmHg*1atm/760mmHg= 0.033 atm Therefore 25 mm Hg is equal to 0.033 atm Because the conversion factor is smaller than 1, and 25 is a lot smaller than the denominator 760 as well, it makes sense that the answer is a lot smaller than 1 and 0.033 atm seems reasonable.

Gas Laws (there are 9 questions)

1. 1. For each of the following, indicate which gas law is described:
   1. a. As volume increases, pressure decreases: Boyle’s Law
   2. b. As temperature increases, volume increases: Charles’s Law
   3. c. As temperature increases, pressure increases: Gay-Lussac’s law
   4. 2. If a balloon is squeezed, what happens to the pressure within the balloon? The pressure in the balloon will increase, as the volume of the balloon will decrease. This can be explained due to Boyle’s law which states, that when one goes up, the other goes down.
   5. 3. What happens to the pressure inside your tires if the temperature decreases? When the temperature decreases the pressure in your tire decreases due to Gay-Lussac’s law in which when one factor decreases, the other decreases as well.
   6. 4. When a balloon is taken from a room at 25oC to the outside at 125oC, what happens to the volume of the balloon? The volume of the balloon will decrease, due to Charles’s law in which when the temperature increase, volume decreases.

Solve the following problems. Show your work for each problem. Include correct units. Enclose your answers in a box. Make sure you determine which gas law you should use before you attempt the problem.

1. 5. What is the new volume of a 87 L sample of gas at 89oC and 107 kPa when it is heated to 95 oC and the pressure increases to 112 kPa? Gas law you will use: Combined Gas Law Work and solution:

Given: T1 = 89oC + (273K/1C)
P1 = 107kPa V1 = 87L T2 = 95oC + (273K/1C) P2 = 112kPa Required: V2 Analyze: I will need to use Combined Gas law to solve the problem and find V2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: V2 = (P1 x T2 x V1)/(T1 x P2)
V2 = (107kPax 368K x 87L)/(362K x 112kPa) (The kPa and K will cancel out)
V2 = 84.49L
Paraphrase:
Therefore the new volume will be at 84.49L.
Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

1. 6. What is the new volume of a 1.9 mL container at 76o C when the temperature is adjusted to 283 oC? (Assume pressure is constant) Gas law you will use: Charles’s law Work and solution:

Given: V1 = 0.0019L T1 = 76oC T­2 = 283 oC Required: V2 Analyze: I will need to use Charles’s law to solve the problem and find V2. The formula needed is V1/ T1 = V2/ T2. Solve: V2 = (V1 x T2)/ T1. To convert from Celsius to Kelvin, I need to use the formula K = C +273. V2 = (0.0019L x (283 oC+273K/1C))/ (76oC+273K/1C) (The C and K will cancel out)
V2 = 2.94mL
Paraphrase: Therefore the new volume will be 2.94mL, if the pressure is constant.
Justify: I know the answer I got is correct, due to the fact if the temperature increases, then the volume will have to decrease to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Charles’s law, when one goes up, the other must go up, which occurs here.

1. 7. A container of gas at 75 oC and 101 kPa increases in pressure to 200 kPa. What is the new temperature of the gas? (assume that volume is constant) Gas law you will use: Gay-Lussac’s law Work and solution:

Given: T1 = 75oC + (273K/1C)
P1 = 101kPa P2 = 200kPa Required: T2 Analyze: I will need to use Gay-Lussac’s law to solve the problem and find T2. The formula needed is P1/ T1 = P2/ T2. Solve: T2 = (T1 x P2)/P1
T2 = (348K x 200kPa)/(101kPa) (The kPa will cancel out)
T2 = 689.11K
Paraphrase:
Therefore temperature of the gas is 681.11K, if the volume is constant.
Justify: I know the answer I got is correct, due to the fact if the pressure increases, then the temperature will have to increase to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Gay-Lussac’s law, when one goes up, the other must go up, which occurs here, as the pressure is almost double so the temperature should also be doubled.

1. 8. What is the new volume when a 65.0 mL container at standard pressure (101.3 kPa) is expanded until the new pressure is 204.3 kPa? (assume the temperature is constant) Gas law you will use: Boyle’s law

Given: V1 = 65mL
P1 = 101.3kPa P2 = 204.3kPa Required: V2 Analyze: I will need to use Boyle’s law to solve the problem and find V2. The formula needed is P1 x V1 = P2 x V2. Solve: V2 = (P1 x V1)/ P2
V2 = (101.3kPa x 65mL)/ 204.3kPa (The kPa will cancel out)
V2 = 32.23mL
Paraphrase:
Therefore the volume of the container will be at 32.23mL, if the temperature is constant.
Justify: I know the answer I got is correct, due to the fact if the pressure increases, then the volume will have to decrease to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Boyle’s law, when one goes up, the other must go down, which occurs here.

1. 9. What is the new pressure of a 15 L sample of gas at 85oC and 101 kPa when it is heated to 125oC and the volume increases to 17.5L? Gas law you will use: Combined Gas Law Work and solution:

Given:T1 = 85 oC + (273K/1C)
P1 = 101kPa V1 = 15L T2 = 125oC + (273K/1C) V2 = 17.5L Required: P2 Analyze: I will need to use Combined Gas law to solve the problem and find P2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: P2 = (P1 x T2 x V1)/(T1 x V2)
P2 = (101kPax 398K x 15L)/(358K x 17.5L) (The L and K will cancel out)
P2 =96.24 kPa
Paraphrase:
Therefore the new pressure will be at 96.24kPa.
Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.
Ideal Gas Law (there are 8 Questions)

1. 1. Under what conditions does a gas behave as ideal? The gas behaves ideal when the gas conforms to the gas laws, the particles exhibit no attractive or repulsive forces, and also when the volume of the particles is assumed to be negligible compared to the total volume occupied by the gas.
2. 2. Under what conditions does a gas not behave as ideal? The conditions that a gas does not behave as ideal is when they have high pressures or low temperatures.
   1. a. Why does this occur? It is because under these conditions the gases tend to condense.
   2. 3. How many moles of gas does it take to occupy 12 liters at a pressure of 20.3 kPa and a temperature of 398 K?

Given: P= 20.3kPa, V=12L, T =398K, R = 8.31LkPa/(Kmol)
Required: n = ?
Analyze: Use the formula n=P x V/ R x T to find n
Solve:
n=P x V/ R x T
n= (20.3kPa) x (12L) / 8.31 LkPa/(Kmol) x 398K = 0.073 mol gas
Paraphrase: It takes 0.073 moles of gas.
Justify: This makes sense because the moles of air is small, therefore the grams will be very small too.

1. 4. If I have a 5.5 liter container that holds 1.2 moles of gas at a temperature of 1250 C, what is the pressure inside the container?

Given: V=5.5L, T =125°C, R = 8.31LkPa/(Kmol), n = 1.2 mol gas
Required: P = ?
Analyze: Use the formula PV=nRT
Solve:
0°C = 273K
K of 125°C= 273K + 125°C
= 398K
P = nRT/V
P= (1.2 mol gas) x (8.31LkPa/(Kmol)) x (398K) / (5.5L)
P= 721.6kPa
Paraphrase: The pressure inside the container is 721.6kPa.
Justify: This makes sense because the volume is small but the temperature and moles are of large value, and according to the formula volume is indirectly proportional to the pressure, and since the volume is small, this has to mean that the pressure has to be is high.

1. 5. It is not safe to put aerosol canisters in a campfire, because the pressure inside the canisters gets very high and they can explode. If I have a 1.0 liter canister that holds 5 moles of gas, and the campfire temperature is 12000 C, what is the pressure inside the canister?

Given: V=1.0L, T =1200°C, R = 8.31LkPa/(Kmol), n = 5 mol gas
Required: P = ?
Analyze: Use the formula PV=nRT
Solve:
0°C = 273K
K of 1200°C= 273K + 1200°C
= 1473K
P = nRT/V
P= (5 mol gas) x (8.31LkPa/(Kmol)) x (1473K) / (1L)
P= 61203.15kPa
Paraphrase: The pressure inside the canister is 61203.15kPa.
Justify: This makes sense because the pressure being 61203.15kPa is very high and it indeed has the ability to explode the canister.

1. 6. How many moles of gas are in a 25 liter scuba canister if the temperature of the canister is 330 K and the pressure is 2300 kPa?

Given: P= 2300kPa, V=25L, T =330K, R = 8.31LkPa/(Kmol),
Required: n = ?
Analyze: Use the formula n=P x V/ R x T to find n
Solve:
n=P x V/ R x T
n= (2300kPa) x (25L) / 8.31 LkPa/(Kmol) x 330K = 20.96 mol gas
Paraphrase: It takes 20.96 moles of gas.
Justify: Since the pressure and volume are pretty high, then we can expect more moles of gas to be present.

1. 7. I have a balloon that can hold 103 liters of air. If I blow up this balloon with 2 moles of oxygen gas at a pressure of 109 kPa, what is the temperature of the balloon?

Given: P= 109kPa, V=103L, n = 2 mol O2, R = 8.31LkPa/(Kmol)
Required: T = ?
Analyze: Use the formula T=P x V/ R x n to find n
Solve:
T=P x V/ R x n
T= (109kPa) x (103L) / 8.31 LkPa/(Kmol) x 2 mol O2 = 657.5K
Paraphrase: the temperature of the balloon is 657.5K.
Justify: This makes sense because the pressure of the volume is high, and with high pressure we know that the temperature is going to be high too.

1. 8. I have a balloon that can hold 0.5 L of air. If I blow up this balloon with 25 grams of oxygen gas at a pressure of 101.3 kPa, what is the temperature of the balloon?

Given: P= 101.3kPa, V=0.5L, R = 8.31LkPa/(Kmol), 25g of oxygen
Required: T = ?
Analyze: First find the moles of Oxygen gas, then use the formula T=P x V/ R x n to find n
Solve:
n O2 gas = 25g O2 gas x 1 mol O2 / 2(16g) O2 gas
n O2 gas = 0.78 mol O2 gas
T=P x V/ R x n
T= (101.3kPa) x (0.5L) / 8.31 LkPa/(Kmol) x 0.78 mol O2 = 7.8K
Paraphrase: the temperature of the balloon is 7.8K.
Justify: This makes sense because the mol of oxygen gas is pretty low, and so is the volume, therefore the temperature is low too.
Partial pressures, effusion, and diffusion (there are 7 questions)

1. 1. Compare and contrast diffusion and effusion: Diffusion is when molecules remain in one container, and move within it, while for effusion the gas escapes the container. Gases which have lower molar mass, can diffuse and effuse a lot more faster than gases with a higher molar mass.
2. 2. Do gases with higher or lower molar masses diffuse faster? Gases with a lower molar mass diffuse faster.
3. 3. Do gases diffuse from high to low concentration or from low to high concentration? Gas diffuse from high concentration to low concentration.
4. A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three gases in the tank are 35 atm of O2, 5 atm of N2, and 25 atm of He, what is the total pressure inside of the tank?

Given:P1 = 35atm
P2 = 5atm P3 = 25atm Required: PT Analyze: I will need to use Dalton’s law of partial pressures to solve for PT. The formula needed is PT = P1 + P2 + P3. Solve: PT = P1 + P2 + P3
PT = 35atm+ 5atm + 25atm
PT =65atm
Paraphrase:
Therefore the total pressure is 65atm.
Justify: I know the answer I got is correct, as when I add up all the pressures I get the total pressure which is bigger than all the pressure values from the 3 other gases.

1. Blast furnaces give off many unpleasant and unhealthy gases. If the total air pressure is 0.99 atm, the partial pressure of carbon dioxide is 0.05 atm, and the partial pressure of hydrogen sulfide is 0.02 atm, what is the partial pressure of the remaining air?

Given: PT = 0.99atm
P1 = 0.05atm P2 = 0.02atm Required: P3 Analyze: I will need to use Dalton’s law of partial pressures to solve for P3. The formula needed is PT = P1 + P2 + P3. Solve: P3 = PT – (P1 + P2)
PT = 0.99atm– (0.05atm + 0.02atm)
PT =0.92atm
Paraphrase:
Therefore the partial pressure for the remaining air is 0.92atm.
Justify: I know the answer I got is correct, as when I add up all the pressures I get the total pressure again.

1. If the air from problem 5 contains 22% oxygen, what is the partial pressure of oxygen near a blast furnace?

Given: P3 = 0.92atm
Required: (O2)Analyze: I will need to use Dalton’s law of partial pressures to solve for P3. The formula needed is PT = P1 + P2 + P3. Solve: O2 = P3 x (22%)
O2 = 0.20atm
O2 =0.20atm
Paraphrase:
Therefore the partial pressure for the O2 is 0.20atm.
Justify: I know the answer I got is correct, as when I use the remaining air from the last question and get 22% of it, the answer is 0.22atm.

1. 7. Compare the effusion rates of NH3 gas and HCl gas. (Show your work!)

Given:
Molar mass NH3­­=17g
Molar mass HCl­­=36.45g
Required: Rate of effusion
Analyze: I will need to use Graham’s law of effusion to solve for rate of effusion. The formula needed is (rateA/rateB) = (molar mass B/molar mass A)-2(same thing as square root). Solve: (36.45g/17g)-2
rateA= rateB(1.46)
Paraphrase:
So the NH3 will effuse 1.46 times faster than HCl.
Justify: So the NH3 will effuse 1.46 times faster than HCl because NH3 is a lot lighter than HCl.


BONUS

8. If I place 3 moles of N2 and 4 moles of O2 in a 27 L container at a temperature of 298 K, what will the pressure of the resulting mixture of gases be?
Moles N2 and O2 = 3 mol + 4 mol =7 mol
Note we can add molN2 and molO2 as they are in a mixture and since there really are no forces of attraction between gases anyways their values can be safely combined and they can be looked at as one gas in our calculations.
Volume = 27 L
Temperature = 298 K
R = 8.31 LkPa/(Kmol)
P = ? kPa
P=nRT/V
= 7 mol* 8.31LkPa/K mol *298 K /27L
= 642.01 kPa
The total pressure in the container is 642.02 kPa
This makes sense because there are a lot of gas particles but the container is only 35 L so it makes sense that the pressure will be high as the particles won't have much space to move around freely and keep colliding with each other, thus increasing the pressure.

Gas Law Study Guide (there are 31 questions)
Remember to study the goals from your notes, read over your notes, read the book, study your previous worksheets, and complete the study guide.
Match the following descriptions with the correct term from this word list: Boyle’s law, Charles’ law, Dalton’s law, Gay-Lussac’s law, Graham’s Law, Ideal Gas Law.

1. For a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. Boyle’s Law
2. The volume of a fixed mass of gas is directly proportional to its Kelvin temperature, if the pressure is kept constant. Charle’s Law
3. pressure of a gas is directly proportional to its Kelvin temperature if the volume is kept constant. _Gay- Lussac’s law
4. P´ V = n ´ R ´ T Ideal Gas Law
5. At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. Dalton’s Law
6. The rate at which a gas will effuse is inversely proportional to the square root of the gas’s molar mass. Graham’s law

Match the following definitions with the correct term from this word list: compressibility, diffusion, effusion, partial pressure

1. 7. a measure of how much the volume of matter decreases under pressure compressibility
2. 8. the pressure exerted by a gas in a mixture partial pressure
3. 9. the escape of gas through a small hole in a container effusion
4. 10. tendency of molecules to move to regions of lower concentration diffusion
5. 11. Why is it easier to compress a gas than a liquid or solid? The space between gas particles are much more than that of a liquid or solid, therefore when under pressure, the gas particles have the ability to move closer together.
6. 12. Why does the pressure inside a container of gas increase if more gas is added to the container? Adding more gas particles to a container is similar to compressing the gas particles, because when compressing, more particles are being present in a smaller volume. Similarly, adding gas particles will cause more gas particles to be present within a given volume thus increasing the pressure.
7. 13. What happens to the temperature of a gas when it is compressed? When the gas is compressed the pressure increase because more gas is being compressed into a smaller volume, and the gas particles move faster. And with increased pressure the temperature increases too.
8. 14. As the temperature of the gas in a balloon decreases, which of the following occurs? The pressure will decrease and the movement of the gas particles will become slower.
9. 15. What happens to the pressure of a gas inside a container if the temperature of the gas decreases? If the temperature of the gas decreases, the pressure will decrease too because the movement of the gas particles is slower than before.
10. 16. If a balloon is heated, what happens to the pressure of the air inside the balloon if the volume remains constant? In this case the pressure increases because the temperature increases but the volume is the same, and we know that with the increase in temperature, there is an increases in speed that will cause an increase in pressure.
11. 17. If a balloon is heated, what happens to the volume of the air in the balloon if the pressure is constant? In this case the volume will increase because this is the only way that the pressure will remain constant.
12. 18.An ideal gas CANNOT be
    1. a. condensed c. heated
    2. b. cooled d. compressed
    3. 19. Under what conditions of temperature and pressure is the behavior of real gases most like that of ideal gases: The conditions of the temperature and pressure is the behavior of real gases most similar to that of the ideal gases when they have low pressures and high temperatures.
    4. 20. Which of the following gases will effuse most rapidly? Hydrogen, nitrogen, oxygen, or helium. Why? Hydrogen because it has the lowest molar mass, and the lower mass the faster an gas will effuse.

For each of the following, show your work, include correct units and box or highlight your answers. Hint: figure out which gas law you need to use first.

1. 
   

The volume of a gas is 25L at 349.0 kPa pressure. What will the volume be when the pressure is reduced to 50.0 kPa, assuming the temperature remains constant?
Given: V1 = 25L
P1 = 349.0kPa P2 = 50kPa Required: V2 Analyze: I will need to use Boyle’s law to solve the problem and find V2. The formula needed is P1 x V1 = P2 x V2. Solve: V2 = (P1 x V1)/ P2
V2 = (349kPa x 25L)/ 50kPa (The kPa will cancel out)
V2 = 174.5L
Paraphrase:
Therefore the volume of the container will be at 174.5L, if the temperature is constant.
Justify: I know the answer I got is correct, due to the fact if the pressure increases, then the volume will have to decrease to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Boyle’s law, when one goes up, the other must go down, which occurs here.

1. A gas has a volume of 55 mL at a temperature of –5.0C. What volume will the gas occupy at 39.0C?

Given: V1 = 55mL
T1 = -5.0oC T­2 = 39.0 oC Required: V2 Analyze: I will need to use Charles’s law to solve the problem and find V2. The formula needed is V1/ T1 = V2/ T2. Solve: V2 = (V1 x T2)/ T1. To convert from Celsius to Kelvin, I need to use the formula K = C +273.
V2 = (55mL x (39.0 oC+273K/1C))/ (-5.0oC+273K/1C) (The C and K will cancel out)
V2 = 64.03mL
Paraphrase:
Therefore the new volume will be 64.03mL, if the pressure is constant.
Justify: I know the answer I got is correct, due to the fact if the temperature increases, then the volume will have to decrease to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Charles’s law, when one goes up, the other must go up, which occurs here.

1. A rigid container of O has a pressure of 987 kPa at a temperature of 713 K. What is the pressure at 273 K?

Given: T1 = 713K
P1 = 987kPa T2 = 273K Required: P2 Analyze: I will need to use Gay-Lussac’s law to solve the problem and find P2. The formula needed is P1/ T1 = P2/ T2. Solve: P2 = (P1 x T2)/T1
P2 = (987kPa x 273K)/(713K) (The kPa will cancel out)
P2 = 377.91kPa
Paraphrase:
Therefore pressure of the O2 is 340.91K, if the volume is constant.
Justify: I know the answer I got is correct, due to the fact if the temperature decreases, then the pressure will have to decrease to maintain the ration of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Gay-Lussac’s law, when one goes down, the other must go down, which occurs here.

1. A gas occupies a volume of 567 mL at 35.0C and 99 kPa. What is the volume of the gas at 273 K and 101.3 kPa (STP)?

Given: T1 = 35oC + (273K/1C)
P1 = 99kPa V1 = 567mL T2 = 273K P2 = 101.3kPa Required: V2 Analyze: I will need to use Combined Gas law to solve the problem and find V2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: V2 = (P1 x T2 x V1)/(T1 x P2)
V2 = (99kPax 273K x 567mL)/(308K x 101.3kPa) (The kPa and K will cancel out)
V2 = 491.16mL
Paraphrase:
Therefore the new volume will be at 491.16L.
Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

1. How many moles of N are in a flask with a volume of 0.870 L at a pressure of 300.0 kPa and a temperature of 300.0 K?

Given: P= 300kPa, V=0.870L, T =300K, R = 8.31LkPa/(Kmol),
Required: n = ?
Analyze: Use the formula n=P x V/ R x T to find n
Solve:
n=P x V/ R x T
n= (300kPa) x (0.870L) / 8.31 LkPa/(Kmol) x 300K = 0.10 mol N2
Paraphrase: It takes 0.10 mol of N2.
Justify: Since the volume is small, and the temperature is a large value, this causes the moles to be smaller.

1. What is the pressure exerted by 87 g of O in a 2.0-L container at 302.0C?

Given: V=2L, T =302°C, R = 8.31LkPa/(Kmol), n = 87g of O
Required: P = ?
Analyze: Use the formula PV=nRT
Solve:
0°C = 273K
K of 302°C= 273K + 302°C
= 575K
n O2 gas = 87g O2 gas x 1 mol O2 / 2(16g) O2 gas
n O2 gas = 2.718 mol O2 gas
P = nRT/V
P= (2.718 mol O2) x (8.31LkPa/(Kmol)) x (575K) / (2L)
P= 6483.64kPa
Paraphrase: The pressure inside the container is 6483.64kPa.
Justify: This makes sense because the pressure being 6483.64kPa is very high is due to the volume being small and the temperature and moles being so high.

1. A mixture of gases at a total pressure of 234 kPa contains N, CO, and O. The partial pressure of the CO is 124 kPa and the partial pressure of the N is 48 kPa. What is the partial pressure of the O?

Given: PT = 234kPa
P1 = 124kPa P2 = 48kPa Required: P3 Analyze: I will need to use Dalton’s law of partial pressures to solve for P3 (O2). The formula needed is PT = P1 + P2 + P3. Solve: P3 = PT – (P1 + P2)
PT = 234kPa – (124kPa + 48kPa)
PT =62kPa
Paraphrase:
Therefore the partial pressure for the remaining air is 62kPa.
Justify: I know the answer I got is correct, as when I add up all the pressures I get the total pressure again.

1. Use Graham’s law to calculate how much slower bromine gas, Br2 , will effuse than chlorine gas, Cl, will.

Given:
Molar mass Br2=159.8g
Molar mass Cl2­­=70.9g
Required: Rate of effusion
Analyze: I will need to use Graham’s law of effusion to solve for rate of effusion. The formula needed is (rateA/rateB) = (molar mass B/molar mass A)-2(same thing as square root). Solve: (159.8g/70.9g)-2
rateA= rateB(1.5)
Paraphrase:
So the Br2 will effuse 1.5 times slower than Cl2.
Justify: This is correct as the Cl2 is a lot lighter than Br2. .

1. How does the air pressure in a balloon change when the balloon is squeezed? Explain why this change occurs. The air pressure in a balloon increases when it has been squeezed because the of Boyle’s law which states that as the volume decreases, the pressure increases. Since squeezing a balloon causes the balloon to loose volume its pressure will increase.
2. How does the pressure of an enclosed gas in a rigid container change when the gas is heated? Explain why this change occurs. The pressure of the enclosed gas in a rigid container will increase because of Gay-Lussac’s law which states that as temperature rises pressure rises, this is because the particles will move much faster as the temperature increases, causing them to collide with the walls which increases the pressure.
3. Explain how pumping air into a bicycle tire increases the pressure within the tire. When air is pumped in the tire more and more gas particles enter the tire and as there are more particles but the same amount of volume the particles collide more often and exert a greater force and this causes the pressure to increase. The pressure keeps increasing until the tires can take the pressure and once they can’t, the tires burst and the air escapes.

GAS LAWS PROBLEMS

Please answer these questions.

Kinetic Theory and Pressure conversions:

For questions 1-5, please answer in full sentences.

1. If you inflate a raft, what will happen to the pressure of the raft as you add more gas?Why does this occur?



When more gas is added to the raft, the pressure inside the raft will increase because according to the kinetic theory, the increase in particles will cause the volume to increase, thus increasing the pressure. And if the pressure exceeds the strength of the raft, then it will burst open.



2. If you heat up an aerosol can, what will happen to the can?Why does this occur?



When the can is heated, the gas particles inside it will start to gain more kinetic energy and move faster. And when the gas particles move faster they strike the walls of the can with more energy. And eventually with the can would burst because with the increase in temperature, the pressure increases too.



3. When you sit on an exercise ball, what will happen to the pressure inside the ball? Why does this occur?



When you sit on an exercise ball, you increase the pressure inside the ball because when you sit, you decrease the volume of the ball. And the decrease in volume will cause an increase in pressure.

1. Use kinetic molecular theory to explain
   1. Why evaporation causes a decrease in a liquid’s temperature
   2. Why gases are more easily compressed than solids and liquids

Evaporation is caused when a particle gains enough energy to escape the intermolecular forces of attraction. When this happens the particle takes this energy along with itself and this energy is gained from the liquid itself. So every time a particle evaporates it take some energy from the liquid with it and this decreases the liquid’s average kinetic energy and thus the temperature.

b. The intermolecular forces are too weak in gases and there is a lot more space between gas particles than solids and liquids. This means that when gases are compressed, the space between particles can be reduced easily as the particles can easily relocate (unlike solids and liquids whose particles cannot move around easily and don’t have space between them) when the space between particles is reduced.

1. A small sample of gas is released in a corner of the room and starts to diffuse to the other side. If the room pressure is increased, will the gas diffuse faster, slower, or at the same speed? Explain.

If the room pressure is increased there will more collisions between particles and they will gain more kinetic energy so their velocity will increase and thus they will diffuse faster.

1. For each of the following, convert from the units given to the desired units using dimensional analysis. SHOW YOUR WORK if you wish to receive credit. Include all units and box or highlight your answers.
   1. 202.6 kPa = ? atm

Pressure=202.6 kPa

Pressure=?atm

For converting kPa à atm, the appropriate conversion factor is 1 atm/ 101.3 kPa

pressure= 202.6 kPa*1atm/101.3kPa=2.0 atm

Therefore 202.6 kPa is equal to 2.0 atm

The answer 2 atm makes sense because at 1 atm the pressure is 101.3 kPa and since 202.6 is exactly double 101.3 therefore the pressure in atm would be double as well

1. 560 kPa = ? mmHg

Pressure=560 kPa

Pressure=?mmHg

For converting kPa àmm Hg, the appropriate conversion factor is 760 mm Hg/ 101.3 kPa

pressure= 560 kPa*760mmHg/101.3kPa=4201.38 mm Hg

Therefore 560 kPa is equal to 4210.38 mm Hg

Because at 101.3 kPa the pressure is 760 mm Hg and since 560 is six times 101.3 it would make sense that the pressure in mm Hg would be almost 6 times 760

1. 5 atm = ? kPa

Pressure=5 atm

Pressure=?kPa

For converting atm à kPa, the appropriate conversion factor is 101.3 kPa/1 atm

pressure= 5atm*101.3kPa/1atm= 506.5 kPa

Therefore 5 atm is equal to 121.56 kPa

Because at 1 atm the pressure is equal to 101.3 and 5 atm is exactly 5 times 1 atm, the answer would be 5 times 101.3 kPa which is 506.5 kPa

1. 3 atm = ? mmHg

Pressure=202.6 kPa

Pressure=?atm

For converting atm à mmHg, the appropriate conversion factor is 760mmHg/1atm

pressure= 3 *760mmHg/1atm= 2280 mm Hg

Therefore 3 atm is equal to 2280 mm Hg

Because at 1 atm the pressure is equal to 760 mm Hg and 3 atm is exactly 3 times 1 atm, the answer would be 3 times 760 mm Hg which is 2280 mm Hg

1. 830 mmHg = ? kPa

Pressure=830 mmHg

Pressure=?kPa

For converting mm Hg à kPa, the appropriate conversion factor is 760 mm Hg/101.3 kPa

pressure= 830 mm Hg*101.3kPa/760 mm Hg= 110.63 kPa

Therefore 830 mm Hg is equal to 110.63 kPa

760 mm Hg is equivalent to 101.3 kPa, and since 830 is only a little greater In value than 760mm, it makes sense that the pressure in kPa will only be a little greater than 101.3 kPa and thus the answer 110.63 kPa seems reasonable

1. 43 mmHg = ? atm

Pressure=43 mmHg

Pressure=?atm

For converting mm Hg à atm, the appropriate conversion factor is 1 atm/ 760 mmHg

pressure= 43mmHg*1atm/760mmHg= 0.057 atm

Therefore 43 mm Hg is equal to 0.057 atm

Because the conversion factor is smaller than 1, and 43 is a lot smaller than the denominator 760 as well, it makes sense that the answer is a lot smaller than 1 and 0.057 atm seems reasonable.





Boyle’s Law



1. What is the new volume when a 125 mL container at 120.0 kPa is expanded until the pressure is 60.0 kPa? (assume the temperature is constant)

Given: V1 = 125mL

P1 = 120kPa P2 = 60kPa Required: V2 Analyze: I will need to use Boyle’s law to solve the problem and find V2. The formula needed is P1 x V1 = P2 x V2. Solve: V2 = (P1 x V1)/ P2

V2 = (120kPa x 125mL)/ 60kPa (The kPa will cancel out)

V2 = 250mL

Paraphrase:

Therefore the volume of the container will be at 250mL, if the temperature is constant.

Justify: I know the answer I got is correct, due to the fact if the pressure decreases, then the volume will have to increase to maintain the ratio of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Boyle’s law, when one goes up, the other must go down, which occurs here.



2. What is the new pressure if a 100.0 L container at standard pressure (101.3 kPa) is compressed until the volume is 50.0 L? (assume the temperature is constant)

Given: V1 = 100L

P1 = 101.3kPa V2 = 50L Required: P2 Analyze: I will need to use Boyle’s law to solve the problem and find P2. The formula needed is P1 x V1 = P2 x V2. Solve: P2 = (P1 x V1)/ V2

P2 = (101.3kPa x 100L)/ 50L (The L will cancel out)

P2 = 202.6kPa

Paraphrase:

Therefore the pressure of the container will be at 202.6kPa, if the temperature is constant.

Justify: I know the answer I got is correct, due to the fact if the volume decreases, then the pressure will have to increase to maintain the ratio of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Boyle’s law, when one goes up, the other must go down, which occurs here.



Charles’s Law



3. What is the new volume of a 10.0 mL container at 0.00 K when the temperature is adjusted to 283 K? (Assume pressure is constant)

Given: V1 = 10mL T1 = 0K T­2 = 283K Required: V2 Analyze: I will need to use Charles’s law to solve the problem and find V2. The formula needed is V1/ T1 = V2/ T2. Solve: V2 = (V1 x T2)/ T1.

V2 = (10mL x 283K)/ (0K)

V2 = undefined

Paraphrase: Therefore there is no new volume, if the pressure is constant.

Justify: I know the answer I got is correct, due to the fact the temperature is at absolute zero, as such there won’t be any volume.



4. A 50.0 mL container is at 273 K. After the volume is adjusted to 100.0 mL, what is the new temperature? (Assume pressure is constant)

Given: V1 = 50mL T1 = 273K V2 = 100ml Required: T2 Analyze: I will need to use Charles’s law to solve the problem and find T2. The formula needed is V1/ T1 = V2/ T2. Solve: T2 = (T1 x V2)/ V1.

T2 = (273K x 100mL)/ (50mL)

T2 = 546K

Paraphrase: Therefore the new temperature is 546K, if the pressure is constant.

Justify: I know the answer I got is correct, V2 is double that of V1, as such T2 should be double T1 to maintain the same ratio which it is.



Gay-Lussac’s Law



5. What is the new pressure of a set volume of gas at 101 kPa when it is heated from 295 K to 400K? (assume that volume is constant)

Given: T1 = 295K

P1 = 101kPa T2 = 400K Required: P2 Analyze: I will need to use Gay-Lussac’s law to solve the problem and find P2. The formula needed is P1/ T1 = P2/ T2. Solve: P2 = (P1 x T2)/T1

P2 = (101kPa x 400K)/(295K) (The K will cancel out)

P2 = 136.95kPa

Paraphrase:

Therefore pressure of the new pressureis 136.95kPa, if the volume is constant.

Justify: I know the answer I got is correct, due to the fact if the temperature increases, then the pressure will have to increases to maintain the ratio of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Gay-Lussac’s law, when one goes up, the other must up, which occurs here.



6. A container of gas at 325K and 500 kPa decreases in pressure to 50 kPa. What is the new temperature of the gas? (assume that volume is constant)

Given: T1 = 325K

P1 = 500kPa P2 = 50kPa Required: T2 Analyze: I will need to use Gay-Lussac’s law to solve the problem and find T2. The formula needed is P1/ T1 = P2/ T2. Solve: T2 = (T1 x P2)/P1

T2 = (325K x 50kPa)/(500kPa) (The kPa will cancel out)

T2 = 32.5K

Paraphrase:

The new temperature of the gas is 32.5K.

Justify: I know the answer I got is correct, due to the fact the original pressure is 10 times as much as the end pressure, as such the temperature will be 10 times as small as the original temperature which it is.



Combined Gas Law



7. What is the new pressure of a 10.5 L sample of gas at 350 K and 101 kPa when it is heated to 375 K and the volume increases to 11.5L?

Given:T1 = 350K

P1 = 101kPa V1 = 10.5L T2 = 375K V2 = 11.5L Required: P2 Analyze: I will need to use Combined Gas law to solve the problem and find P2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: P2 = (P1 x T2 x V1)/(T1 x V2)

P2 = (101kPax 375K x 10.5L)/(350K x 11.5L) (The L and K will cancel out)

P2 = 98.8kPa

Paraphrase:

Therefore the new pressure will be at 98.8kPa.

Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.



8. What is the new volume of a 15 L sample of gas at 365 K and 107 kPa when it is heated to 395 K and the pressure increases to 112 kPa?

Given: T1 = 365K

P1 = 107kPa V1 = 15L T2 = 395K P2 = 112kPa Required: V2 Analyze: I will need to use Combined Gas law to solve the problem and find V2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: V2 = (P1 x T2 x V1)/(T1 x P2)

V2 = (107kPax 395K x 15L)/(365K x 112kPa) (The kPa and K will cancel out)

V2 = 15.51L

Paraphrase:

Therefore the new volume will be at 15.51L.

Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.



9. What is the new temperature of a 4 L sample of gas at 375 K and 102 kPa when it is expanded to 9 L and has a pressure of 95 kPa?

Given: T1 = 375K

P1 = 102kPa V1 = 4L V2 = 9L P2 = 95kPa Required: T2 Analyze: I will need to use Combined Gas law to solve the problem and find T2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: T2 = (P2 x T1 x V2)/(V1 x P1)

T2 = (95kPax 375K x 9L)/(4L x 102kPa) (The kPa and K will cancel out)

V2 = 785.85K

Paraphrase:

Therefore the new temperature will be at 785.85K.

Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

More Problems:

1. 1. What is the new volume of a 15 L sample of gas at 67oC and 107 kPa when it is heated to 95 oC and the pressure increases to 112 kPa?

Given: T1 = 340K

P1 = 107kPa V1 = 15L T2 = 368K P2 = 112kPa Required: V2 Analyze: I will need to use Combined Gas law to solve the problem and find V2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: V2 = (P1 x T2 x V1)/(T1 x P2)

V2 = (107kPax 368K x 15L)/(340K x 112kPa) (The kPa and K will cancel out)

V2 = 15.51L

Paraphrase:

Therefore the new volume will be at 15.51L.

Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

1. 2. What is the new volume of a 10.0 mL container at 42o C when the temperature is adjusted to 283 oC? (Assume pressure is constant)

Given: V1 = 10mL T1 = 313K T­2 = 556K Required: V2 Analyze: I will need to use Charles’s law to solve the problem and find V2. The formula needed is V1/ T1 = V2/ T2. Solve: V2 = (V1 x T2)/ T1.

V2 = (10mL x 556K)/ (313K)

V2 = 17.76mL

Paraphrase: Therefore the new volume is 17.76mL, if the pressure is constant.

Justify: I know the answer I got is correct, due to the new temperature is almost double the old temperature and this is also see with the volume, showing that the ratio is 1: 1.

1. 3. A container of gas at 25 oC and 101 kPa increases in pressure to 200 kPa. What is the new temperature of the gas? (assume that volume is constant)

Given: T1 = 298K

P1 = 101kPa P2 = 200kPa Required: T2 Analyze: I will need to use Gay-Lussac’s law to solve the problem and find T2. The formula needed is P1/ T1 = P2/ T2. Solve: T2 = (T1 x P2)/P1

T2 = (298K x 200kPa)/(101kPa) (The kPa will cancel out)

T2 = 590.1K

Paraphrase:

The new temperature of the gas is 590.1K.

Justify: I know the answer I got is correct, due to the fact the original pressure is 2 times less as the end pressure, as such the temperature will be 2 times as more as the original temperature which it is.

1. 4. What is the new volume when a 50.0 mL container at standard pressure (101.3 kPa) is expanded until the new pressure is 2.5 atm? (assume the temperature is constant)

Given: V1 = 50mL

P1 = 101.3kPa P2 = 253.31kPa Required: V2 Analyze: I will need to use Boyle’s law to solve the problem and find V2. The formula needed is P1 x V1 = P2 x V2. Solve: V2 = (P1 x V1)/ P2

V2 = (101.3kPa x 50mL)/ 253.31kPa (The kPa will cancel out)

V2 = 19.93mL

Paraphrase:

Therefore the volume of the container will be at 19.93mL, if the temperature is constant.

Justify: I know the answer I got is correct, due to the fact if the pressure increases, then the volume will have to decrease to maintain the ratio of 1 : 1 between the starting gas and the gas in the container. This also is correct because due to Boyle’s law, when one goes up, the other must go down, which occurs here.

1. 5. What is the new pressure of a 10.5 L sample of gas at 75oC and 101 kPa when it is heated to 125oC and the volume increases to 11.5L?

Given:T1 = 348K

P1 = 101kPa V1 = 10.5L T2 = 398K V2 = 11.5L Required: P2 Analyze: I will need to use Combined Gas law to solve the problem and find P2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: P2 = (P1 x T2 x V1)/(T1 x V2)

P2 = (101kPax 398K x 10.5L)/(348K x 11.5L) (The L and K will cancel out)

P2 = 105.47kPa

Paraphrase:

Therefore the new pressure will be at 105.47kPa.

Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

1. 6. Explain why gases at the same temperature and pressure have different densities (in g/L)?

Gases at the same temperature and pressure have different densities because; all gases have different molar mass. Some gases like hydrogen gas only weight around 2 grams per mol, while other like oxygen gas weight around 32g per mol. This then translates into different densities. As such even though gases have the same temperature and pressure they might not have the same density.

Remember to convert from oC to K whenever temperature is given in oC.

1) How many moles of gas does it take to occupy 120 liters at a pressure of 202.3 kPa and a temperature of 340 K?

Given: P= 202.3kPa, V=120L, T =340K, R = 8.31LkPa/(Kmol),

Required: n = ?

Analyze: Use the formula n=P x V/ R x T to find n

Solve:

n=P x V/ R x T

n= (202.3kPa) x (120L) / 8.31 LkPa/(Kmol) x 340K = 8.59 mol gas

Paraphrase: It takes 8.59 mol of gas

Justify: This makes sense because since the volume and pressure is high, and according to the formula we know that both the volume and pressure are directly proportional to each other, therefore we can expect the mol to be high.

2) If I have a 50 liter container that holds 45 moles of gas at a temperature of 2000 C, what is the pressure inside the container?

Given: V=50L, T =200°C, R = 8.31LkPa/(Kmol), n = 45 of gas

Required: P = ?

Analyze: Use the formula PV=nRT

Solve:

0°C = 273K

K of 200°C= 273K + 200°C

= 473K

P = nRT/V

P= (45 mol gas) x (8.31LkPa/(Kmol)) x (473K) / (50L)

P= 3537.56kPa

Paraphrase: The pressure inside the canister is 3537.56kPa.

Justify: This makes sense because the volume being small and the temperature and moles being so high. And according to the formula we see that the pressure is directly proportional to the moles, temperature but indirectly proportional to the volume.



3) It is not safe to put aerosol canisters in a campfire, because the pressure inside the canisters gets very high and they can explode. If I have a 1.0 liter canister that holds 2 moles of gas, and the campfire temperature is 14000 C, what is the pressure inside the canister?

Given: V=1L, T =1400°C, R = 8.31LkPa/(Kmol), n = 2 moles of gas

Required: P = ?

Analyze: Use the formula PV=nRT

Solve:

0°C = 273K

K of 1400°C= 273K + 1400°C

= 1673K

P = nRT/V

P= (2 mol gas) x (8.31LkPa/(Kmol)) x (1673K) / (1L)

P= 27805.26kPa

Paraphrase: The pressure inside the canister is 27805.26kPa.

Justify: This makes sense because a very high pressure will be needed to explode a canister, and I think that 27805.26kPa is a very high pressure considering the fact that the temperature is very high, and we know that as the temperature increases the pressure increases simultaneously.

4) How many moles of gas are in a 30 liter scuba canister if the temperature of the canister is 300 K and the pressure is 2026 kPa?

Given: P= 2026kPa, V=30L, T =300K, R = 8.31LkPa/(Kmol),

Required: n = ?

Analyze: Use the formula n=P x V/ R x T to find n

Solve:

n=P x V/ R x T

n= (2026kPa) x (30L) / 8.31 LkPa/(Kmol) x 300K = 24.38 mol gas

Paraphrase: It takes 24.38 mol of gas

Justify: This makes sense because pressure is high, and the temperature is not as much high, and since the moles are directly proportional to the pressure and indirectly proportional to the temperature, we can expect a high amount moles of gas.

5) I have a balloon that can hold 100 liters of air. If I blow up this balloon with 3 moles of oxygen gas at a pressure of 101.3 kPa, what is the temperature of the balloon?

Given: P= 101.3kPa, V=100L, n = 3 mol O2, R = 8.31LkPa/(Kmol)

Required: T = ?

Analyze: Use the formula T=P x V/ R x n to find T

Solve:

T=P x V/ R x n

T= (101.3kPa) x (100L) / 8.31 LkPa/(Kmol) x 3 mol O2 = 406.3K

Paraphrase: the temperature of the balloon is 406.3K.

Justify: This makes sense because the pressure is high, and with high pressure we know that the temperature is going to be high too.

6) If I have a 67 L container that holds 5 grams of nitrogen gas at a temperature of 25oC, what is the pressure inside the container?

Given: V=67L, T =25°C, R = 8.31LkPa/(Kmol), n = 5g of nitrogen gas

Required: P = ?

Analyze: Use the formula PV=nRT

Solve:

0°C = 273K

K of 25°C= 273K + 25°C

= 298K

n N2 gas = 5g N2 x 1 mol N2 / 2(14g) N2 gas

n N2 gas = 0.178 mol N2 gas

P = nRT/V

P= (0.178 mol N2) x (8.31LkPa/(Kmol)) x (298K) / (67L)

P= 6.6kPa

Paraphrase: The pressure inside the container is 6.6kPa.

Justify: This makes sense because the temperature is low and so is the moles, and according to the formula we know that the pressure is directly proportional to the temperature and moles, and indirectly proportional to the volume. And since the temperature and moles are low, we can expect the pressure to be low as well.



7) If I have a 32 L container that holds 5.4 x 1023 molecules of oxygen gas at a pressure of 101.3 kPa, what is the temperature inside the container?

Given: P= 101.3kPa, V=32L, 5.4 x 1023 O2 molecules, R = 8.31LkPa/(Kmol)

Required: T = ?

Analyze: Use the formula T=P x V/ R x n to find T

Solve:

n= 5.4 x 1023 molecules of O2 x 1 mol O2 / 6.02 x 1023 molecules of O2

n=0.89 mol O2

T=P x V/ R x n

T= (101.3kPa) x (32L) / 8.31 LkPa/(Kmol) x 0.89 mol O2 = 438.29K

Paraphrase: the temperature of the balloon is 438.29K.

Justify: This makes sense because the pressure is high, and with a high pressure we know that the temperature is going to be high too.



Partial Pressures and Graham’s Law Practice:

1. A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three gases in the tank are 3 atm of O2, 5 atm of N2, and 7 atm of He, what is the total pressure inside of the tank?

P 02 = 3 atm

P N2=5 atm

PHe=7 atm

Ptotal=? atm

Ptotal= P 02+ PHe+P N2

= 3atm+8atm+7atm

= 18atm

Therefore the total pressure inside the gas tank is 18 atm

The answer seems reasonable as all the values are smaller than it and it is the sum

1. Blast furnaces give off many unpleasant and unhealthy gases. If the total air pressure is 1.45 atm, the partial pressure of carbon dioxide is 0.43 atm, and the partial pressure of hydrogen sulfide is 0.23 atm, what is the partial pressure of the remaining air?

P 02 = 6.6 kPa and P N2=23.0 kPa, what is P CO2

PH2=0.23

Ptotal=1.45atm

PCO2= 0.43atm

Ptotal= PN2+ P02+ PCO2

Premaining air=?atm

Premaining air = Ptotal -PCO2- PH2

= 1.45atm-0.43atm-0.23atm

= 0.79atm

The partial pressure of the remaining gas is 0.79 atm

The partial pressure of the remaining gas should be greater than that of carbon dioxide and hydrogen because their values are really small but should be smaller than the sum and thus 0.79atm seems reasonable.

1. If the air from problem 2 contains 22% oxygen, what is the partial pressure of oxygen near a blast furnace?

Ptotal=1.45atm

Percent O2 = 0.22%

P O2 = 22*1.45atm*1/100=0.319atm

1. Compare the effusion rates of O2 (molar mass, 32.0 g/mol) and N2 (molar mass, 28.0 g/mol).

Molar mass O2­­=32g/mol

Molar mass N2­­=28g/mol



So from the same hole nitrogen will escape 1.069 times faster than oxygen. This makes sense because at the same temperature both gases possess the same kinetic energy but the mass of oxygen is higher, and thus the velocity of nitrogen will be higher as it has less mass.

1. If I place 3 moles of N2 and 4 moles of O2 in a 27 L container at a temperature of 298 K, what will the pressure of the resulting mixture of gases be?

Moles N2 and O2 = 3 mol + 4 mol =7 mol

Note we can add molN2 and molO2 as they are in a mixture and since there really are no forces of attraction between gases anyways their values can be safely combined and they can be looked at as one gas in our calculations.

Volume = 27 L

Temperature = 298 K

R = 8.31 LkPa/(Kmol)

P = ? kPa

P=nRT/V

= 7 mol* 8.31LkPa/K mol *298 K /27L

= 642.01 kPa

The total pressure in the container is 642.02 kPa

This makes sense because there are a lot of gas particles but the container is only 27 L so it makes sense that the pressure will be high as the particles won't have much space to move around freely and keep colliding with each other, thus increasing the pressure.

Mixed Gas Problems:

1. 1. If 690.0 mL of oxygen is collected over water at 26.0 ˚C and 725 mmHg, what is the dry volume of this oxygen sample at 52.0 ˚C and 106.6 kPa?

Given: T1 = 299K

P1 = 96.66kPa V1 = 690mL T2 = 325K P2 = 106.6kPa Required: V2 Analyze: I will need to use Combined Gas law to solve the problem and find V2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: V2 = (P1 x T2 x V1)/(T1 x P2)

V2 = (96.66kPax 325K x 690mL)/(299K x 106.6kPa) (The kPa and K will cancel out)

V2 = 680.07mL

Paraphrase:

Therefore the new dry volume will be at 680.07mL.

Justify: I know the answer I got is correct, as when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

1. 2. If 400.0 mL of hydrogen gas is collected over water at 18.0˚C and a total pressure of 98.6 kPa, what would be the dry volume of this hydrogen sample at STP?

Given: T1 = 291K

P1 = 98.6kPa V1 = 400mL T2 = 273.15K P2 = 101.33kPa Required: V2 Analyze: I will need to use Combined Gas law to solve the problem and find V2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: V2 = (P1 x T2 x V1)/(T1 x P2)

V2 = (98.6kPax 273.15K x 400mL)/(291K x 101.33kPa) (The kPa and K will cancel out)

V2 = 365.35mL

Paraphrase:

Therefore the new dry volume will be at 365.35mL.

Justify: I know the answer I got is correct, because if the temperature and pressure only deviate a little from the original, then the volume must also be only a bit different than the original volume, which it is. Also when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

1. 3. A 6.12 L sample of xenon gas was collected over water at 2.00 X 105 Pa and 80.0˚C. What would be the gas volume of pure xenon at SATP?

Given: T1 = 353K

P1 = 200kPa V1 = 6.12L( At SATP conditions Temperature is 298.15K, and pressure is at 101.325pKa.) T2 = 298.15K P2 = 101.33kPa Required: V2 Analyze: I will need to use Combined Gas law to solve the problem and find V2. The formula needed is (P1 x V1)/ T1 = (P2 x V2)/ T2. Solve: V2 = (P1 x T2 x V1)/(T1 x P2)

V2 = (200kPax 298.15K x 6.12L)/(353K x 101.33kPa) (The kPa and K will cancel out)

V2 = 10.2L

Paraphrase:

Therefore the new dry volume will be at 10.2L.

Justify: I know the answer I got is correct, as the temperature increases, while the pressure decreases, this tells us that the volume will have to increase which it does. Also when I use the original formula and solve both sides of the equation I get the ration of 1:1 which means my answer is correct.

1. 4. How many moles of gas are contained in 890.0 mL at 21.0˚C and 750.0 mmHg?

Given: P= 750 mmHg, V=890mL, T =21.0°C, R = 8.31LkPa/(Kmol)

Required: n = ?

Analyze: Use the formula n=P x V/ R x T to find n

Solve:

P=750 mm Hg*101.3kPa/760 mm Hg= 99.9kPa

P=99.9kPa

V= 890mL x (1L / 1000mL)

V= 0.89L

0°C = 273K

K of 21°C= 273K + 21°C

= 294K

n=P x V/ R x T

n= (99.9kPa) x (0.89L) / 8.31 LkPa/(Kmol) x 294K = 0.036 mol gas

Paraphrase: It takes 0.036 mol of gas

Justify: This makes sense because pressure and the volume is low and we know according to the formula that the moles is directly proportional to the pressure and the volume, therefore it is not a surprise that the mole of gas is low.

1. 5. What volume will 20.0 g of Argon gas occupy at STP?

Given: mass = 20g at STP molar mass of Argon = 39.95g/mol

Required: VAnalyze: I will need to use a bit of Stoichiometry. Solve:

V = 20g x ( 1 mol/ 39.95g) x (22.4L/1 mol)

V = 11.21L

Paraphrase:

Therefore the volume of Argon will be at 11.21L.

Justify: I know the answer I got is correct, as the mass of Argon gas given is around half the mass of 1 mol of Argon, as such the volume should be half the volume of a gas at normal STP, which it is.

1. 6. What volume would 32.0 g NO2 gas occupy at 3.12 atm and 18.0˚C?

Given: P= 3.12 atm, T =18.0°C, R = 8.31LkPa/(Kmol), 32g NO2 gas

Analyze: Use the formula PV=nRT

Solve:

0°C = 273K

K of 18°C= 273K + 18°C

= 291K

P= 3.12atm*101.3kPa/1atm= 316.05kPa

n= 32g NO2 x 1 mol NO2 / (14g) + 2(16g)

n= 0.69 mol NO2

V = nRT/P

V= (0.69mol NO2) x (8.31LkPa/(Kmol)) x (291K) / (316.05kPa)

V= 5.29L

Paraphrase: The volume is 5.29L.

Justify: This makes sense because the according o the formula we know that the temperature and moles are directly proportional to the volume, and these factors being small cause the volume to be small as well.

1. 7. Calculate the molar mass of a gas if 35.44 g of the gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at 35.5˚C.

Given: P= 60.0atm, V=7.5L, T =35.5.0°C, R = 8.31LkPa/(Kmol) , 35.44g

Required: M= ?

Analyze: Use the formula n=P x V/ R x T to find n

Solve:

P=60atm*101.3kPa/1atm= 6078kPa

P=6078kPa

0°C = 273K

K of351°C= 273K + 35.5°C

= 308.5K

n=P x V/ R x T

n= (6078kPa) x (7.5L) / 8.31 LkPa/(Kmol) x 308.5K = 17.8 mol gas

M=17.8 mol gas x (35.44g of gas / 1 mol gas)

M=631.1g of gas

Paraphrase: The molar mass is 631.1g of gas

Justify: This does make sense because the moles are of a high value, therefore the molar mass too should be of a high value.

1. 8. Determine the number of moles of Krypton contained in a 3.25 L tank at 5.88 X 105 Pa and 25.5˚C. If the gas was oxygen instead of krypton, will the answer be the same?

Given: P= 5.88 x 105 Pa, V=3.25L, T =25.5°C, R = 8.31LkPa/(Kmol) ,

Required: n= ?

Analyze: Use the formula n=P x V/ R x T to find n

Solve:

P=5.88 x 105 P x 1kPa / 1000Pa

P=588kPa

0°C = 273K

K of25.5°C= 273K + 25.5°C

= 298.5K

n=P x V/ R x T

n= (588kPa) x (3.25L) / 8.31 LkPa/(Kmol) x 298.5K = 0.77 mol krypton

n= 0.77 mol gas

Paraphrase: Theerfore it is 0.77 of krypton. Even if the krypton was replaced by oxygen, the mol would be the same but the molar mass would be different.

Justify: This does make sense because the moles doesn’t show the characteristics of an element but the molar mass does.

1. 9. Determine the mass of carbon dioxide in a 450.6 mL flask at 1.80 atm and –50.5˚C. Determine the mass of oxygen that would be present in the same container under the same conditions.

Given: P= 1.8atm, V=450.6mL, T =-50.5°C, R = 8.31LkPa/(Kmol) ,

Required: M= ?

Analyze: Use the formula n=P x V/ R x T to find n

Solve:

P=1.8atm x 101.3kPa/1atm

P=182.34kPa

V= 450.6mL x (1L / 1000mL)

V= 0.4506L

0°C = 273K

K of-50.5°C= 273K + (-50.5°C)

= 222.5K

n=P x V/ R x T

n= (182.34kPa) x (0.4506L) / 8.31 LkPa/(Kmol) x 222.5K = 0.044 mol CO2

n= 0.044 mol CO2

M of CO2=0.044 mol CO2 x (12g) + 2(16g) CO2 / 1 mol CO2

M of CO2=1.936 g of CO2

M of O2 = 0.044 mol CO2 x (2 mol O2 / 1 mol CO2) x (16g O2/ 1 mol O2)

M of O2 = 1.408g O2





Paraphrase: The molar mass of carbon dioxide is 1.936g and the molar mass of oxygen is 1.408g

Justify: This make sense because the mole ration is 1:2, therefore there are two oxygen atoms and oxygen element has a greater mass than that of carbon.

1. 10. At what temperature will 0.654 mol neon gas occupy 12.30 L at 197.5 kPa?

Given: P= 197.5kPa, V=12.30L, n = 0.654 mol neon, R = 8.31LkPa/(Kmol)

Required: T = ?

Analyze: Use the formula T=P x V/ R x n to find n

Solve:

T=P x V/ R x n

T= (197.5kPa) x (12.3L) / 8.31 LkPa/(Kmol) x 0.654 mol neon = 446.9K

Paraphrase: At the temperature of 446.9K.

Justify: This makes sense because the pressure of the volume is high, and with high pressure we know that the temperature is going to be high too.

1. 11. A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 303.9 kPa, and 3040 mm Hg respectively. What is the total pressure of this sample in kPa, atm, and mmHg?

Given: P1 = 2atm, P2 = 303.9kPa, P3 = 3040mmHg

Required: Pt Analyze: I will need to use Dalton’s law Solve:

PT = P1 + P2 + P3

PT = 2atm + (303.9 kPa * (1 atm/101.325 kPa)) + ( 3040mmHg * (1 atm/760mmHg))

PT= 9atm

PT= 9atm x (101.325 kPa /1 atm)

PT= 911.93kPa

PT= 9atm x (760mmHg /1 atm)

PT= 6840mmHg

Paraphrase:

Therefore the total pressure can be expressed as 9atm, 911.93kPa, and 6840mmHg.

Justify: I know the answer I got is correct, as when u subtract each value from the total, you can get back the original pressure values.

1. 12. A tank contains 480.0 g of oxygen and 80.0 g of helium at a total pressure of 7.00 mmHg.

What are the partial pressures of oxygen and helium?

Mass O2=480 g

Mass He=80 g

nO2=480g O2*1 mol O2/16g O2

= 30 mol O2

nHe= 80g He*1 mol He/4 g He

= 20 mol He

Total gas particles = 20 mol + 30 mol = 50 mol

% composition O2=30*100%/50=60% O2

% composition He=20*100%/50=40% He

Partial pressure of O2=60/100*7mmHg=4.2mmHg

Partial pressure of He=40/100*7mmHg=2.8mmHg

Therefore the partial pressure of oxygen in the tank is 4.2 mmHg and the partial pressure of helium in the tank is 2.8 mmHg

The answer makes sense as the partial pressure 4.2mmHg and 2.8mmHg, add up to 7mmHg



Gas Stoichiometry:

1. Pentane, C5H12(l), burns completely to form carbon dioxide and water.

a) Write the balanced chemical equation for this reaction.

C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(g)

b) What volume of O2(g) at STP is required to produce 70.0 L of CO2(g) at STP?

Given:

V CO2=70L

V O2=?L

T=0°C=273.15K

P=101.3kPa

R = 8.31 LkPa/(Kmol)

PV=nRT

n=PV/(R*T)

n CO2=101.3kPa*70L/(8.31 LkPa/(Kmol)*273.15K)

= 3.12mol

n O2 = 3.12molCO2*8mol O2/5 mol CO2

= 4.99mol

V O2=nRT/P

= 4.99mol*8.31 LkPa/(Kmol)*273.15K/101.3kPa

= 111.81L

Therefore 111.81L of oxygen gas is required to produce 70.0L of carbon dioxide at STP

Because there is 70L CO2 and 5 moles of it are produced when 8 moles of oxygen are used it would mean that the amount of oxygen used was more as both gases are experiencing the same pressure and temperature and their particles take up the same amount of space. The answer 111.81L makes sense because 8 moles of oxygen are used up in order to produce 5 moles of carbon dioxide.

c) What mass of H2O(l) is made when the combustion of C5H12(l) gives 106 L of CO2(g) at SATP?

V CO2=106L

mass H2O=?L

T=0°C=273.15K

P=101.3kPa

R = 8.31 LkPa/(Kmol)

PV=nRT

n=PV/(R*T)

n CO2=101.3kPa*106L/(8.31 LkPa/(Kmol)*273.15K)

= 4.73mol

Mass H2O = 4.73molCO2*6mol H2O/5 mol CO2*(1+1+16)g H2O/1mol H2O

= 102.17g

Therefore 102.17g water is produced along when 70L carbon dioxide is produced at STP when C5H12(l) is burnt

1. Given the equation C3H8(g) + O2(g) → CO2(g) + H2O(g)

a) Balance the equation

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

b) If 50 g of C3H8 is burned in excess O2 , what volume of CO2 gas can be collected at 30∧C and 90 kPa?

T=30∧C=303.15K

P=90 kPa

R = 8.31 LkPa/(Kmol)

Mass C3H8=50g


= 3.4mol CO2

V CO2 =nRT/P

= 3.4mol*8.31 LkPa/(Kmol)*303.15K/90kPa

= 95.17L

Therefore 95.17L carbon dioxide is produced when 50 g of C3H8 is burned in excess O2 at 30∧C and 90 kPa

1. 3. 2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g)

a) What volume of O2 at SATP is required for the reaction of 1.46 g of ZnS?

Mass ZnS=1.46g


nSO2=0.015mol

SATP

T=298.15 K

P=100kPa

R = 8.31 LkPa/(Kmol)

V O2 =nRT/P

= 0.015mol*8.31 LkPa/(Kmol)*298.15K/100kPa

=0.37L

Therefore 0.37L oxygen is required to react with 1.46g of ZnS


b) What volume of SO2 at SATP will be produced from the reaction in a)?

Mass ZnS=1.46g


nSO2=0.226mol

SATP

T=298.15 K

P=100kPa

R = 8.31 LkPa/(Kmol)

V SO2 =nRT/P

= 0.226mol*8.31 LkPa/(Kmol)*298.15K/100kPa

=5.599L

Therefore 5.599L sulfur dioxide is produced from the reaction of 1.46 g of ZnS

Because only 1.46g of ZnS was used in the reaction and the molar ratio is 2:2 it makes sense that the answer will be small too, and thus 5.599L seems reasonable.

1. 4. Given the equation (NH4)2SO4(aq) + 2KOH(aq) → 2NH3(g) + K2SO4(aq) + 2H2O(l)

a) Calculate the volume of ammonia gas, measured at 23∧C and 64 kPa, that could be produced from 264 g of ammonium sulfate and 280.0 g of potassium hydroxide.

Mass (NH4)2SO4 =264g


= 4 mol

T=296.15 K

P=64kPa

R = 8.31 LkPa/(Kmol)

V NH3 =nRT/P

= 4mol*8.31 LkPa/(Kmol)*296.15K/64kPa

=153.81L

Therefore 153.81L NH3 is produced from the reaction of 264g (NH4)2SO4

Mass KOH =280g


n NH3=5mol

T=296.15 K

P=64kPa

R = 8.31 LkPa/(Kmol)

V NH3 =nRT/P

= 5mol*8.31 LkPa/(Kmol)*296.15K/64kPa

=192.26L

Therefore 192.26L NH3 is produced from the reaction of 280g KOH

Because (NH4)2SO4 has a greater molar mass and a smaller quanitity but 1:2 molar ratio it makes sense that less volume of NH3 is formed than 280g of KOH which is lighter. It would make sense that the volume of ammonia gas would be high because the pressure is really low and thus it would indicate that the gas particles have a lot of space between them indicating a higher volume.

1. 5. Given the equation 2KClO3(s) → 2KCl (s) + 3O2(g)

What volume of a gas can be produced by the decomposition of 122.6 g of potassium chlorate measured under the following conditions?

Mass KClO3=122.6g

nO2=122.6g*1mol KClO3/122.55g KClO3*3 mol O2/2 mol KClO3

= 1.5mol O2

a) at STP

T=0°C=273.15K

P=101.3kPa

R = 8.31 LkPa/(Kmol)

Mass KClO3=122.6g

VO2=nRT/P

=1.5mol*8.31 LkPa/(Kmol)*273.15K/101.3kPa

= 33.61L

b) at SATP

T=298.15 K

P=100kPa

R = 8.31 LkPa/(Kmol)

V O2=nRT/P

= 1.5mol*8.31 LkPa/(Kmol)*298.15K/100kPa

=37.16L

Therefore 33.61L of gas is produced at STP conditions while 37.16L of gas is produced at SATP conditions

Because the pressure and temperature are higher at SATP than STP it would make sense that the volume of gas produced at SATP conditions would be greater as the gas particles will have greater kinetic energy as the temperature is higher.